Question

The following reaction has $\Delta H$ as $-25kcal$.

$C{H}_4\left(g\right)+C{l}_2\left(g\right)\to C{H}_{3}Cl\left(g\right)+HCl\left(g\right)$

Bond	Bond Energy kCal
${\epsilon }_{C-Cl}$	84
${\epsilon }_{H-Cl}$	109
${\epsilon }_{C-H}$	$x$
${\epsilon }_{Cl-Cl}$	$y$
$x:y=9:5$

From the given data, what is the bond enthalpy of $Cl-Cl$ bond?

• A. $70kcal$
• B. $80kcal$
• C. $67.75kcal$
• D. $108kcal$

Answer 1

Answer: (D)

$108kcal$

$\Delta H_{rxn}=B.E{.}_R-B.E{.}_P$

$-25=(4X+y)-(3X+109+84)$ or $-25+193=X+Y$

$168=X+\frac{5X}{9}$ or $168=\frac{14X}{9}$

$\frac{168\times 9}{14}=X$ or $X=108Kcal$